# Give sequences such that $Y_n\neq 0$ $\forall n \in \Bbb N$ and lim($X_n - Y_n) = 0$ but lim $\frac {X_n}{Y_n} \neq 1$

Give sequences $X_n$, $Y_n$ such that:

1) $Y_n\neq 0$ $\forall n \in \Bbb N$ and lim($X_n - Y_n) = 0$ but lim $\frac {X_n}{Y_n} \neq 1$

2)$Y_n\neq 0$ $\forall n \in \Bbb N$ and lim($X_n - Y_n) = \infty$ and lim $\frac {X_n}{Y_n} = 1$

3)$Y_n\neq 0$ $\forall n \in \Bbb N$ and lim($\frac{X_n - Y_n}{n^2}) = \infty$ and lim $\frac {X_n}{Y_n} = 1$

I have to solve a lot of excercises like these but I do not how to do it, just for the 1st one: I think that if lim($X_n - Y_n) = 0$ the $X_n$ and $Y_n$ are to close, so $\frac {X_n}{Y_n} \approx 1$ and the limit must be 1, or not?

Is it possible to give sequences like these? Any ideas?

• Hint: For the first, consider $X_n = \frac{2}{n}$, $Y_n = \frac{1}{n}$. The others can be solved with $X_n$ a polynomial of $n$. – stochasticboy321 Oct 21 '15 at 1:23

HINT: For the first one try $X_n=\frac1n$ and $Y_n=\frac1{2n}$. For the second, remember that if $p(x)$ and $q(x)$ are any quadratic polynomials with leading coefficient $1$, then $\lim_{x\to\infty}\frac{p(x)}{q(x)}=1$. The third one just needs a slightly fancier version of the same idea.
For the first one, $x_n = \frac1{n}$ and $y_n = \frac1{n^2}$ has $x_n/y_n \to \infty$.
For the second, $x_n = n^2+n$ and $y_n = n^2$.
For the third, $x_n = 4^n+3^n$ and $y_n=4^n$ satisfies $\frac{x_n-y_n}{y_n} \to 1$ and $\frac{x_n-y_n}{n^k} \to \infty$ for any $k > 0$.